Density is a material property, so the second cube will have the same density as the first.
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.
Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²
We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m
Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
= </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
= 39.6 N/C
Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
Answer:
0.098 N
Explanation:
From the question,
Spring scale reading = W-U............... Equation 1
Where W = weight of the cube, U = upthrust.
W = mg
Where m = mass of the cube, g = acceleration due to gravity.
Given: m = 11 g = 0.011 kg, g = 9.8 m/s².
W = 0.011(9.8)
W = 0.1078 N.
From Archimedes principle,
Upthrust = weight of water displaced.
U = (Density of water×volume of metal cube)×acceleration due to gravity.
U = (D×V)g
Given: D = 1000 kg/m², V = 1 cm³ = (1/1000000) = 1×10⁻⁶ m³, g - 9.8 m/s²
U = 1000(9.8)(10⁻⁶)
U = 0.0098 N.
Substitute the value of W and U into equation 1
Reading of the spring scale = 0.1078-0.0098
Reading of the spring scale = 0.098 N
B) 48.0 m/s
We can actually start to solve the problem from B for simplicity.
The motion of the rock is a uniformly accelerated motion (free fall), so we can find the final speed using the following suvat equation
where
is the final velocity
is the initial velocity (positive since we take downward as positive direction)
is the acceleration of gravity
s = 110 m is the vertical displacement
Solving for v, we find the final velocity (and so, the speed of the rock at impact):
A) 3.67 s
Now we can find the time of flight of the rock by using the following suvat equation
where
is the final velocity at the moment of impact
is the initial velocity
is the acceleration of gravity
t is the time it takes for the rock to reach the ground
And solving for t, we find
